PROPAGATION OF STRESS WAVES IN ELASTIC SOLIDS
Ion Al. Cr˘aciun
Abstract. The propagation of waves in elastic media under dynamic loads (stress waves) are investigated. The nature of deformation, stress, stress–strain relations, and equation of motion are some objectives of inves-tigation. General decompositions of elastic waves are studied. Two planar waves in an infinite isotropic elastic medium, and then time–harmonic so-lutions of the wave equations are analyzed. Spherically symmetric waves in three–dimensional space from a point source, radially symmetric waves in a solid infinite cylinder of radius a, and waves propagated over the surface of an elastic body are studied. Finally, particular solutions of the Navier equations are given.
2000 Mathematics Subject Classification: 74B05, 74J05, 74J10, 74J15.
Keywords: elastic waves, time–harmonic solutions, spherically and radi-ally symmetric waves.
1. Introduction
The field equations of physics for a given continuous medium arise from the three conservation equations: conservation of mass, momentum, and energy. For an elastic medium these equations of motion are called theNavier equations. From them, a rich variety of stress waves is obtained. The first two conservation laws are common to all media, while the energy equation defines a particular medium through the equation of state. When we deal with an elastic solid we are concerned with the dynamic variables: stress and strain. The energy equation for an elastic solid yields a relation between the stress and strain. For small–amplitude deformations there is a linear relation between the stress and strain, which is given by Hooke’s law.
2. Notation and Mathematical Preliminaries
Physical quantities are mathematically represented by tensors of various or-ders. The equations describing physical laws are tensor equations. Quantities that are not associated with any special direction and are measured by a sin-gle number are represented by scalars, or tensors of order zero. Tensors of order one are vectors, which represent quantities that are characterized by a direction as well as a magnitude. More complicated physical quantities are represented by tensors of order greater than one.
Throughout this paper light–faced Roman or Greek letters stand for scalars, Roman letters in boldface denote vectors, while lower case Greek letters in boldface denote second–order tensors.
2.1 Indicial notation
A sistem of fixed rectangular Cartesian coordinates is sufficient for the pre-sentation of the theory. In indicial notation, the coordinates axes may be denoted by xj and the unit base vectors by ej, where j = 1,2,3. In the se-quel, subscripts assume the values 1,2,3 unless explicitly otherwise specified. If the components of a vector u are denoted by uj, we have
u=u1e1+u2e2+u3e3. (1) Since summation of the type (1) frequently occur in the mathematical de-scription of the mechanics of a continuum medium, we introduce the summa-tion convensumma-tion, whereby a repeated subscript implies a summation. Equa-tion (1) may be rewritten as
u =ujej. (2)
As another example of the use of the summation convention, the scalar product of the two vectors is expressed as
u·v=ujvj =u1v1+u2v2+u3v3. (3) As opposed to the free index in uj, which may assume any one of the values 1,2,3, the indexj in (2) and (3) is a bound index or a dummy index, which must assume all three values 1,2 and 3
tensor of rank three. A well-known special tensor of rank two is theKronecker delta, whose components are defined as
δij =
(
1 if i=j
0 if i6=j.
A frequently–used special tensor of rank three is the alternating tensor, whose components are defined as follows:
εijk =
+1 if ijk represents an even permutation of 123 0 if any two of theijk indices are equal
−1 if ijk represents an odd permutation of 123.
By the use of the alternating tensor and the summation convention, the components of the cross product h=u×v may be expressed as
hi =εijkujvk. In extended notation the components of h are
h1 =u2v3 −u3v2, h2 =u3v1−u1v3, h3 =u1v2−u2v1. 2.2 Vector operators
Particularly significant in vector calculus is the Hamilton’s vector operator (or nabna) denoted by ∇,which is given by
∇=e1 ∂
∂x1 +e2
∂ ∂x2
+e3
∂ ∂x3
.
When applied to the scalar field ϕ(x1, x2, x3), the vector operator ∇ yields a vector field which is known as the gradientof the scalar field,
gradϕ =∇ϕ= ∂ϕ
∂x1 e1+
∂ϕ ∂x2
e2+
∂ϕ ∂x3
e3.
In indicial notation, partial differentiation is commonly denoted by a comma, and thus
The appearence of a single subscript in ϕ,k indicates that ϕ,k are the components of a tensor of rank one, i. e.., a vector.
In a vector field, denoted by u(x),the components of the vector are func-tions af spatial coordinates. The components are denoted by ui(x1, x2, x3). Assuming that functions ui(x1, x2, x3) are differentiable, the nine partial derivatives ∂ui
∂xj
(x1, x2, x3) can be written in indicial notation as ui,j. It can be shown that ui,j are the components of a second–rank tensor.
When the vector operator ∇operates on a vector in a manner analogous to scalar multiplication, the result is a scalar field, termed thedivergence of the vector field u(x)
divu=∇·u=ui,i.
By taking the cross product of ∇ and u, we obtain a vector termed the
curl ofu,denoted by curlu=∇×u. Ifq=∇×u, the components ofqare
qi =εijkuk,j.
The Laplace operator ∇2 is obtained by taking the divergence of a gra-dient. The Laplacian of a twice differentiable scalar field is another scalar field,
div gradϕ=∇·∇ϕ =∇2ϕ =f,ii.
The Laplacian of a vector field is another vector field denoted by ∇2u ∇2u=∇·∇u=uk,jjek.
2.3 Gauss’ Theorem
One of the most important integral theorem of tensor analysis, known as
Gauss’ theorem, relates a volume integral to a surface integral over the bound-ing surface of the volume. Consider a convex regionB of volumeV,bounded by a surface S which possesses a piecewise continuously turning tangent plane. Such a region is said to be regular. Now let us consider a tensor field
τjkl···p,and let every component of τjkl···p be continuously differentiable inB. Then Gauss’ theorem states
Z
V
τjkl···p,idV = Z
S
niτjkl···pdA, (4)
successively substituted for τjkl···p, and if the three resulting expressions are
added, the result is
Z
V
ui,idV =
Z
S
niuidA. (5)
Equation (5) is the well–known divergence theorem of a vector calculus which states that the integral of the outer normal component of a vector over a closed surface is equal to the integral of the divergence of the vector over volume bounded by the closed surface.
2.4 Notation
The equations governing the linearized theory of elasticity are presented in the following commonly used notation:
position (radius) vector : x(coordinates xi) displacement vector : u(components ui) small strain tensor : ε(componentsεij) stress tensor : τ(componentsτij). 3. Fundamental Concepts in Elasticity
In this section we shall investigate the nature of deformation, strain, stress, stress–strain relations, equations of motion for the stress components, and equations of motion for the displacement.
3.1 Deformations
An elastic material is a deformable, continuous medium which suffers no energy loss when its deformed state returns to the equilibrium state. The deformation of any medium is a purely geometric concept. Since strain is derived from deformation, strain is also a geometrical concept. A continuum is a medium that has a continuous distribution of matter in the sense that its molecular and crystalline structure is neglected. This means that we can define mathematically a differential volume element dV that has the same continuous properties as the material in the large. This concept of a continuum is based on an averaging process, where we take advantage of the large number of molecules in a differential volume element to smear out the effects of individual molecules.
1. The undeformed or equilibrium configuration or state; 2. The deformed configuration.
Any two neighboring points P1, P2 in the body in its undeformed state under a deformation suffer the transformation P1 7→ P1′, P2 7→ P2′. The distance between the two undeformed points changes in the deformed state. If
P′
1P
′
2 < P1P2 that part of the body undergoes a compression; ifP
′
1P
′
2 > P1P2 we have tension. Clearly, if there is no change we have the equilibrium state. We use the Lagrange representation. A point in the undeformed state is given by the coordinates X = (X1, X2, X3), where X is the radius vector. In the deformed state, that point goes into the Eulerian coordinates given by the radius vector x, where x= (x1, x2, x3). The displacement vector u is defined by u = x−X. Since we use the Lagrange representation, we have x=x(X, t),u =u(X, t),and all the dynamic and thermodynamic variables are functions of (X, t)). We have thus defined deformation (compression or tension) in terms of the displacement vector.
3.2 Strain Tensor
For the equilibrium state, letX= (X1, X2, X3) and for the deformed state let x= (x1, x2, x3) be the radius vectors of an arbitrary point of the body. In the equilibrium state a particle occupies the volume elementdVX =dX1dX2dX3. In the deformed state that particle occupies the volume element
dVx=dx1dx2dx3.
A deformation is given by the regular transformation X 7→ x, which has a unique inverse transformation. The relation between these two volume elements is
dVx = det(JX(x))dVa,
where det(JX(x)) is the determinant of JX(x),the Jacobian of the transfor-mationX7→x. JX(·) is also called the mapping functionand is given by the matrix
JX(x) =
∂x1
∂X1
∂x1
∂X2
∂x1
∂X3
∂x2
∂X1
∂x2
∂X2
∂x2
∂X3
∂x3
∂X3
∂x3
∂X2
∂x3
∂X3
The principle of conservation mass is given by
ρxdVx = ρXdVX. The compression ratioR is defined by
R= ρx
ρX
=hdet(JX(x))i
−1
= det(Jx(X)).
We develop the strain tensor in the form of a matrix by considering a curve CX in the body in the deformed state. Under the mapping X 7→ x, this curve maps into the curveCx in the deformed state. The two curves are composed by the same particles. The column matrixdX has the components
dX1, dX2, dX3, while the row matrix is the transpose dX∗. We have dX = (dX1, dX2, dX3)∗, and similarly for dx. Let dsX be an element of arc lenght ofCX and letdsxbe an element of arc length of Cx under the transformation
dX7→dx which is given by
dx = JX(x)dX. (6)
The magnitude of dsX is the square root of the expression (dsX)2 = (dX)
∗
(dX).
Similarly, we have
(dsx)2 = (dx)∗
(dx).
The transpose of equation (6) is (dx)∗
= (dX)∗
(JX(x))∗
. Using this ex-pression and equation (6), we get
(dsx)2 = (dx)∗(dx) = (dX)∗(JX(x))∗JX(x)(dX).
Suppose the transformation X7→xhas the property that for every curve
CX all arc lengths are unchanged in being transformed to the corresponding curveCx.It follows that (dx)
∗
(dx) = (dX)∗
(dX),so that (JX(x))∗
JX((X)) = E, the 3×3 identity matrix. This means that JX(x) is the rotation ma-trix yielding a rigid–body rotation (no deformation), where det(JX(x))>0.
Since the measure of strain can be considered as a deviation from a pure rotation, we may take the expression (JX(x))∗
JX((X)) − E as twice the three–dimensional strain tensor. Thus, we have
ε = 1
2((JX(x))
∗
whereε is the strain tensor. The reason for the factor of 1/2 in equation (7) is seen when we derive the linear approximation ofε.For this approximation we have |(dsx−dsX)/dsX| ≪1,so that
(dsx)2−(ds X)2 (dsX)2
= (dsx−dsX)(dsx+dsX) (dsX)2 ≈ 2dsX(dsx−dsX)
(dsX)2
= 2dsx−dsX
dsX
.
LetεL be the linear strain tensor. We have
dsx−dsX
dsX
= dX
dsX ∗ εL dX dsX ,
where equation (7) is used for the linear strain strain tensor. 3.3 Strain as a Function of Displacement
We now obtain the strain tensor matrix ε as a function of the displacement vector u,where u= (u1, u2, u3).We have u=x−X, the components being
ui = xi−Xi. Instead of calculating the right–hand side of (7) and setting x=u+X to obtain ε, we introduce K=JX(u), the Jacobian ofu, so that
K = ∂u1 ∂X1 ∂u1 ∂X2 ∂u1 ∂X3 ∂u2 ∂X1 ∂u2 ∂X2 ∂u2 ∂X3 ∂u3 ∂X1 ∂u3 ∂X2 ∂u3 ∂X3 .
It follows thatJ =K+E, so that the strain tensor in matrix form becomes
ε = 1
2(K+K
∗
) +K K∗ = εL+εN. (8) The linear part of relation (8) is given by the matrix
εL =
∂u1 ∂X1 1 2 ∂u1 ∂X2
+ ∂u2
∂X1
1
2
∂u1 ∂X3
+ ∂u3
∂X1 1 2 ∂u 2 ∂X1
+ ∂u1
∂X2 ∂u 2 ∂X2 1 2 ∂u 2 ∂X3
+ ∂u3
∂X2 1 2 ∂u3 ∂X1
+ ∂u1
∂X3
1
2
∂u3 ∂X2
+ ∂u2
The diagonal elements of ε represent the pure or normal components of the strain, while the off–diagonal elements are the components of the shear strain. The strain tensor ε is represented by a symmetric matrix. This is true for both the linear and nonlinear parts, that is for both εL and εN.
3.4 Linear Momentum and the Stress Tensor
A basic postulate in the theory of continuous media is that the mechanical action of the material points which are situated on one side of an arbitrary material surface within a body upon those on the other side can be completely accounted for the prescribing a suitable surface traction on this surface. Thus if a surface element has a unit outward normal n we introduce the surface traction t, defining a force per unit area. The surface tractions generally depend on the orientation of n as well as on the location x of the surface element.
Suppose we remove from a body a closed region V +S, where S is the boundary. The surface S is subjected to a distribution of surface tractions t(x, t).Each mass element of the body may be subjected to a body force per unit mass,F(x, t).According to the principle of balance of linear momentum, the instantaneous rate of change of the linear momentum of a body is equal to the resultant external force acting on the body at the particular instant of time. In the linearized theory this leads to the equation
Z
S
tdA+
Z
V
ρFdV =
Z
V
ρ∂
2u
∂t2 . (10)
By means of the tetrahedron argument, equation (10) subsequently leads to the stress tensor τ with componentsτkl,where
tl=τklnk (11)
Equation (11) is the Cauchy stress formula. Physically τkl is the compo-nent in the xi−direction of the traction on the surface with the unit normal ik.
By substitution of tl = τklnk, equation (10) is rewritten in an indicial notation as
Z
S
τklnkdA+
Z
V
ρFldV =
Z
V
ρ∂
2u l
The surface integral can be transformed into a volume integral by Gauss’ theorem, and we obtain
Z
V
∂τkl ∂xk
+ρFl−ρ
∂2u l
∂t2
dV = 0.
Since V may be an arbitrary part of the body it follows that wherever the integral is continuous, we have
∂τkl
∂xk
+ρFl =ρ
∂2u l
∂t2 . (12)
This is Cauchy’s first law of motion and it represents the conservation of linear momentum. These equations of motion are therefore Newton’s equa-tions of motion for a continuum. They are the linearized equaequa-tions of motion, since the nonlinear terms for the particle acceleration, which represent the convective terms, are neglected. Since they are linear, the Lagrangian and Eulerian representations are the same; the difference between these two rep-resentations appears only in the nonlinear terms. The equations of motion are valid for any continuous medium (solid, liquid, gas) since they do not invoke the energy equation that defines that material. Clearly, the equations of motion give an incomplete description of the physical situation, since we have three equations and six components of the stress tensor τ and three components of the displacement vectoru.The additional requisite equations are given from the conservation of energy, which supplies the constitutivwe equations; these we take as Hooke’s law for an isotropic elastic medium. 3.5 Balance of Moment of Momentum
For the linearized theory the principle of moment of momentum states
Z
S
(x×t)dA+
Z
V
(x×F)dV =
Z
V
ρ ∂ ∂t
x×∂u
∂t
dV.
Simplifying the right–hand side and introducing indicial notation, this equa-tion can be written as
Z
S
εklmxltmdA+
Z
V
εklmxlFmdV =
Z
V
ρ εklmxl
∂2u l
Elimination oftm from the surface integral and the use of Gauss’ theorem result in Z
S
εklmxlτkmnkdA=
Z
V
εklm(δlkτkm+xlτkm,k)dV. By virtue of the first law of motion, equation (13) reduces to
Z
V
εklmδlkτkmdV
or
εklmτlm = 0. This result implies that
τlm =τml, i. e., the stress tensor is symmetric.
4. The Navier Equations of Motion for the Displacement
In this section we derive the linear equations of motion for an elastic medium in terms of the components of the displacement vector by using Hooke’s law. These equations are called the Navier equations. In many problems in elasticity it is more convenient to obtain the equations of motion for the displacement vector and then derive the stress field from the definition of strain and Hooke’s law.
4.1 Stress–Strain Relations
In general terms, the linear relation between the components of the stress tensor and the components of the strain tensor is
τij =Cijklεkl, where
Cijkl =Cjikl =Cklij =Cijlk.
of the Cartesian coordinate system in which the components of τij and εij are evaluated. It can be shown that the constantsCijkl may be expressed as
Cijkl=λδijδkl+µ(δikδjl+δilδjk). Hooke’s law then assumes the well–known form
τij =λεkkδij + 2µεij. (14)
Equation (14) contains two elastic constants λ and µ, which are known as Lam´e’s elastic constants.
The trace (sum of the diagonal elements) of the strain matrix Θ is called
dilatation. It is a measure of the relative change in volume of the body due to a compression or dilatation. It is given by
Θ = ε11+ε22+ε33. (15)
Hooke’s law can be also expressed by the set of equations
τij =λδijΘ + 2µεij. (16) Let Φ be the trace of stress matrix so that
Φ =τ11+τ22+τ33. (17)
We have
Φ = (3λ+ 2µ)Θ. (18)
The inverse of equation (16) is
εij =
λ
2µ(3λ+ 2µ)Φδij + 1
2µτij. (19)
The Lam´e constants λ, µ can be expressed in terms of the two elastic con-stantsYoung’s modulus E and Poisson’s ratio σ.It can be shown that
λ = Eσ
(1 +σ)(1−2σ), µ=
E
2(1 +σ)
σ = λ
2(λ+µ), E =
µ(3λ+ 2µ)
λ+µ .
4.2 Equations of Motion for the Displacements
We obtain the linear strain in tensor form from equation (9) by using the x
coordinate system instead of the Lagrangian variables X and the dispalace-ment vector u. The ijth component of the linear strain tensor is εij = ui,j. Since the strain tensor is symmetric, we have
εij = 1
2(ui,j+uj,i). (21)
τkl,k =µ∇2ul+ (λ+µ)(∇·u),l. (22) In this way, the lth equation of motion (12) becomes
µ∇2ui+ (λ+µ)(∇·u),i=ρ(−Fi+ui,tt). (23) The vector form of the Navier equations of motion for the displacement can be obtained from (23) by multiplying with the unit vector ei of the xi axis followed by addition over i.Thus, we have
µ∇2u+ (λ+µ)∇Θ =ρ(−F+utt). (24) Using the well known identity
∇2u =∇(∇·u)−curlΨ=∇(∇·u)−∇×Ψ,
where Ψ = curlu = ∇×u is the curl of displacement vector u, equation (24) becomes
−µ∇×Ψ+ (λ+ 2µ)∇Θ =ρ(−F+u,tt). (25) Ψ is called the rotation vector. Note that the vector form of the Navier equations are independent of the coordinate system. From the form given by (25) we now derive two types of waves: (1) longitudinal waves involving the wave equation for the dilatation, and (2) transverse waves, involving the rotation vector.
(1) We take the divergence of each term of equation (25) and use the fact that ∇·∇×Ψ= 0. We obtain
λ+ 2µ
ρ ∇
Equation (26) is the vector form of the three–dimensional nonhomoge-neous wave equation for Θ,the nonhomogeneous term being the one involving F.We set
λ+ 2µ
ρ =c
2 L.
It will be seen below that cL is the longitudinal wave velocity.
(2) Next we get the equation of motion for Ψby taking the curl of each term in equation (25). We use the fact that ∇×Θ =0 and obtain
µ ρ∇
2Ψ=Ψ
,tt−∇×F.
We set µ
ρ =c
2 T,
where cT is the velocity of the rotational vector, which, we shall shall show, is the velocity of a transverse wave.
5. Propagation of Plane Elastic Waves
In this section we study the plane waves in an infinite isotropic elastic medium. Aplane waveis defined as one whose wave front is a planar surface normal to the direction of the propagating wave. If the wave front is normal to x1 axis, then the displacement vector is a function of coordinate x1 and timet,and all derivatives with respect tox2 andx3 in equation (24) are zero. For simplicity we set F = 0. Since the dilatation becomes Θ = u1,1, we see that the three scalar equations in (24) become
c2
Lu1,11 = u1,tt
c2
Tu2,11 = u2,tt
c2
Tu3,11 = u3,tt.
(27)
that we hve a transverse wave, and cT is indeed the transverse wave veloc-ity. Similarly for the third equation, where u3(x1, t) shows transverse wave propagation with the same velocitycT. From their definitions it is seen that the velocity of longitudinal waves is always greater than that of transverse waves.
6. General decomposition of Elastic Waves
By using the definitions ofcL andcT we can write the vector Navier equation (24) in the form
c2T∇2u+ (c2L−c2T)∇(∇·u) = u,tt, (28) where, again, F =0. We now split equation (28) into two vector equations by decomposing the dispalacement vector as follows:
u=uL+uT. (29)
We shal show thatuLrepresents a longitudinal wave anduT a transverse wave. A longitudinal wave is rotationless, which means that
∇×uL = 0.
It follows from vector analysis that a scalar function of space and time exists such that
uL= gradφ=∇φ=
∂φ ∂x1
e1+
∂φ ∂x2
e2+
∂φ ∂x3
e3, (30)
where φ is called the scalar potential. On the other hand, uT satisfies the equation
divuT = 0. (31)
This clearly means that a transverse wave suffers no change in volume (an equivoluminal wave) but is rotational, so that
u=∇×ψ, (32)
where ψ is the vector potential.
Using equations (30) and (32),equation (29) becomes
Equation (32) tells us that the displacement vector can be decomposed into an irotational vector and an equivoluminal vector.
We now insert equation (29) into (28),take the divergence of each term, and use equation (31). We obtain
∇ ·(c2L∇2uL−uL,tt) = 0.
Since∇×uL =0and∇·( ) = 0,it follows that the terms in parantheses are also zero, yielding
c2L∇2u
L−uL,tt =0. (34)
Equation (34) is the vector wave equation for the displacement repre-senting longitudinal wave (irrotational waves), since the wave velocity is cL. Since uL =∇φ, it is clear that the scalar potential φ also satisfies the wave equation with the same velocity.
Similarly, inserting equation (29) into (28), taking the curl of each term, and using the fact that ∇×uL =0,we obtain
∇×(c2T∇2uT −uT,tt) = 0. Since ∇ ·( ) = 0, it follows that
c2T∇2uT −uT,tt =0. (35)
Equation (35) is the vector wave equation for uT, whose solutions yield transverse, equivoluminal, rotational waves. It follows that the vector poten-tial ψ also staisfies this wave equation.
Sometimes it is easier to solve the wave equations in terms of the scalar and vector potentials. Then the displacement can be obtained from equation (33).
7. Characteristic Surfaces for Planar Waves
The wave front of a planar wave is a plane normal to the direction of wave propagation. Let the normal to the wave front beν = (l, m, n),wherel, m, n
are the direction cosines ofν.Let the radius vectorx=x1e1+x2e2+x3e3 be directed from the origin to aby point on the wave front. The scalar product ν ·xis the projection of x on the wave front.
All solutions of the wave equations (34) and (35) are of the form uL = f(ν·x+ct) +g(ν·x−ct),
where f, g, F and G are arbitrary functions of the indicated arguments which are called the phases of the waves, (ν ·x+ct), and (ν ·x−ct) are called the phases of the waves. The functions f(ν ·x+ct), F(ν · x+ct) represent progressing waves, while g(ν ·x−ct) and G(ν ·x−ct) are the regressing waves. It is also true that there are no solutions that are not of the form given by equations (36). If we set ν ·x−ct = const., we obtain a characteristic surface; this is a planar wave front that progress into the medium in the direction normal to the wave front with a wave velocity equal to c. Setting ν·x+ct= const. yields a characteristic surface that regresses in the direction oposite to that of the normal to the wave front. Inserting u = uL+uT, where uL and uT are given in (36), into equation (24), and setting then x=y=z = 0, yields the following quadratic equation for c2 :
(ρ c2−µ)(ρ c2−λ−2µ) = 0. (37) The roots of equation (37) are c2 = c2
L, c2T. This substantiates the fact that, in any direction ν, there are two planar waves, one longitudinal and the other transverse.
8. Time–Harmonic Solutions of the Reduced Wave Equations
We now investigatetime–harmonicsolutions to the wave equations for longi-tudinal and transverse waves. These are waves whose time–dependent parts are of the form e±iωt
.
We write the displacement vector u in the form u = Re [U(x1, x2, x3)e
±iωt
], (38)
where Re[ ] is the real part of the bracket. In equation (38), u stands for either a longitudinal or a transverse wave. Substituting equation (38) into equations (34) and (35) and factoring out the exponentials yields
∇2U
L+kL2UL = 0, ∇2UT +k2
TUT = 0,
(39)
where kL and kT are the wave numbers for the longitudinal and transverse waves, respectively, and are given by
kL=
ω cL
, kT =
ω cT
Equations (39) are called the reduced wave equations for UL and UT, respectively. The reduced wave equations are of eliptic type. Let U stands forUL and UT.Solutions of the reduced wave equation∇2U+k2U=0are of the form
U=U¯eik(ν·x) ,
where U¯ is a constant vector. Combining this solution of the reduced wave equation with the time–dependent solution given by equation (38) yields the time–harmonic solutions
u= Re [U¯ eik(ν·x±ct)
], (40)
where c=ω/k.
The minus sign in the phase gives a progressing travelling wave and the plus sign a regressing wave, ω is the same for both a longitudinal and trans-verse waveform, while c and k depend on the waveform.
The time–harmonic solutions expressed by equations (40) are special case of the general solution given by (36). Note that we have the same phase of the progressing and regressing waves, given by (ν ·x±ct), so that planes of constant phase yield the traveling wave fronts.
9. Spherically Symmetric Waves
Spherically symmetric waves are produced in three–dimensional space from a point source. For simplicity we consider the wave equation for the generic scalar f(r, t), which may stand for the components ofu,the scalar potential
φ,or the components of the vector potentialψ.Settingr2 =x2
1+x22+x23 and neglecting angular dependence, we get the wave equation for f in spherical coordinates:
c2hf,rr +
2
r
f,r
i
=f,tt. (41)
We get time–harmonic solutions of equation (41) by setting
f(r, t) =g(r)e−iωt
. (42)
Inserting equation (42) into (41) yields
d2g
dr2 +
2
r
dg dr +k
2g = 0, k= ω
This ordinary differential equations for g may be rewritten as
d2(rg)
dr +k
2(rg) = 0. (43)
Note that equation (43) is a second order ordinary differential equation for
rgwhere the first–derivative term is absent. Settingw=rggivesw′′
+k2w= 0, so that the general solution of equation (43) is
g(r) = 1
r)e ±ikr
.
Multiplying this solution by the time–dependent part given by equation (42) yelds time–harmonic solutions of the spherical wave equation as a linear combination of terms of the form
1
re
i(kr−ct)
, 1
re
i(kr+ct) (44)
The first expression of equation (44) represents an outgoing attenuated spherical wave (emanating from the point source), while the second expres-sion represents an incoming wave (from infiniti where the amplitude is zero) going toward the source. Note that there is a singularity at the source. The phase of the wave is (kr ∓ct). Setting the phase (k +ct) equal to a constant and varying time generates an outgoing spherical wave front, and setting (k−ct) equal to a constant generates an incoming wave; these are the characteristic surfasces.
We may obtain a more general solution to the spherical wave equation (41) by writing it as
c2(rf),rr = (rf),tt.
Therefore the general solution for the spherical wave equation is
f(r, t) = 1
rF(r−ct) +
1
rG(r+ct), (45)
10. Curvilinear Orthogonal Coordinates
In this section we digress by investigating the transformations from Cartesian to curvilinear orthogonal coordinates and then specialize to the transforma-tion to cylindrical coordinates.
Let (q1, q2, q3) be coordinates of a point in any system. The Cartesian coordinates x1, x2, x3 will be functions of these coordinates, so that the set of functions
x1 =x1(q1, q2, q3), x2 =x2(q1, q2, q3), x3 =x3(q1, q2, q3), (46) is a regular transformation, which can be written asx=x(q).Consequently, there is the inverse transformation
q1 =q1(x1, x2, x3), q2 =q2(x1, x2, x3), q3 =q3(x1, x2, x3).
The three equationsqi =ci,whereci are constants, represent three families of surfaces whose lines of intersection form three families of curved lines. These lines of intersection will be used as the coordinate lines in our curvilinear coordinate system. Thus, the position of a point in space can be defined by the values of three coordinates q1, q2 and q3. The local coordinate directions at a point are tangent to the three coordinate lines intersecting at the point. The Jacobian matrix of transformation (46) is a 3×3 matrix whose kth column has the elements xk,i. This matrix will be denoted by Jq(x).
The differentialsdxi are now expanded in terms of (dq1, dq2, dq3),so that
dxi =
∂xi
∂q1
dq1+
∂xi
∂q2
dq2 +
∂xi
∂q3
dq3, i= 1,2,3. (47) In the matrix form, equations (47) can be written as
dx∗ =dq∗Jq(x)
∗ ,
where dx∗ is the transpose matrix of the one column matrixdx whose com-ponents are dx1, dx2, dx3. Similarly, for dq
∗
.Let ds be an element of length. It follows that
ds2 = 3
X
i=1 3
X
j=1
where
gij =gji = 3
X
k=1
∂xk
∂qi ·
∂xk
∂qj
(49) are called the metric coefficients.
The quadratic form (48) can be written as
ds2 =dq∗Jq(x)
∗
Jq(x)dq.
If we consider the element of lengthdsi,wherei= 1,2,3,that corresponds to a change from qi to qi+dqi, we have
ds1 =√g11dq1, ds2 =√g22dq2, ds3 =√g33dq3. To shorten the notation we introduce the scale factors
h1 =√g11, h2 =√g22, h3 =√g33, which are in general functions of the coordinates qj.
For an orthogonal system of coordinates the surfaces q1 = const, q2 = const, q3 = const intersect each other at right angles, and allgij, with i6=j, are equal to zero. We choose an orthonormal right–handed basis whose unit vectors e1,e2 and e3 are respectively directed in the sense of increase of the coordinates q1, q2 and q3.The following well-known relations hold
ei·ej =δij,
ei×ej =ek, (50)
where in (50) the indices i, j and k are in cyclic order. A major difference between curvilinear coordinates and Cartesian coordinates is that the coor-dinatesq1, q2 andq3 are not necessarily measured in lengths. This difference manifests itself in the appearence of scale factors in the relation between the infinitesimal displacement vectordxand the infinitesimal variationsdq1, dq2 and dq3, namely,
All equations in Cartesian coordinates which do not involve space deriva-tives and which pertain to properties at a point carry over unchanged into curvilinear coordinates. If space derivatives are involved, however, equations do not directly carry over, since the differential operators such as gradient, divergence, curl, and the Laplacian assume different forms.
We consider first the gradient operator ∇. When applied to a scalarϕ it gives a vector ∇ϕ, with components which we call f1, f2 and f3.Thus
∇ϕ =f1e1+f2e2+f3e3. (52) The increment of ϕ due to a change of position dxis
dϕ=∇ϕ·dx=h1f1dq1 +h2f2dq2+h3f3dq3, (53) where (51) has been used. The increment dϕ can also be written as
dϕ= ∂ϕ
∂q1
dq1+
∂ϕ ∂q2
dq2+
∂ϕ ∂q3
dq3, (54)
whence it can be concluded that ∇= e1
h1
∂ ∂q1
+ e2
h2
∂ ∂q2
+ e3
h3
∂ ∂q3
. (55)
If the operation (55) is applied to the scalar function ϕ, then from (53) and (54) it results that fi in (52) is equal to
1
hi
∂ϕ ∂qi
, so that the gradient of the scalar function ϕ gives
gradϕ =∇ϕ= 1
h1
∂ϕ ∂q1
e1+ 1
h2
∂ϕ ∂q2
e2+ 1
h3
∂ϕ ∂q3
e3. (56) Let us consider an other vector function F which in the q orthogonal system is expressed by
F=F1e1+F2e2 +F3e3. (57) where (F1, F2, F3) are the components of F along the axes of the q system depending ofq1, q2 and q3.
The divergence of Fin this orthogonal system of coordinates is ∇·F= 1
h1h2h3
∂(h2h3F1)
∂q1
+∂(h3h1F2)
∂q2
+∂(h1h2F3)
∂q3
while the curl of the same vector is given by ∇×F = 1
h2h3
∂(h3F3)
∂q2 −
∂(h2F2)
∂q3
e1+
+ 1
h3h1
∂(h1F1)
∂q3 −
∂(h3F3)
∂q1
e2+
+ 1
h1h2
∂(h2F2)
∂q1 −
∂(h1F1)
∂q2
e3.
(59)
Note that the expression (59) of the vector ∇×F can be write in the form of a determinant, namely,
∇×F= 1
h1h2h3
h1e1 h2e2 h3e3
∂ ∂q1 ∂ ∂q2 ∂ ∂q3
h1u1 h2u2 h3u3
. (60)
TheLaplacian operatorof a scalar function can easily be derived by using (58) and (55). Thus
∇2ϕ=∇·∇ϕ= 1
h1h2h3 3 X n=1 ∂ ∂qn
h1h2h3
h2 n ∂ϕ ∂qn . (61)
In cylindrical coordinate we choose
q1 =r, q2 =θ, q3 =z.
The corresponding scale factors and unit base vectors are
h1 = 1, h2 =r, h3 = 1 e1 =er, e2 =eθ, e3 =ez.
(62)
The relations between the cylindrical coordinates (r, θ, z) and the Carte-sian coordinates (x1, x2, x3) are
x1 = r cosθ,
x2 = r sinθ,
x3 = z.
In this case, r is the length of the projection on the plane (x1, x2) of the radius vector,θ is the angle between this projected radius vector and x1 axis measured in the trigonometric sense, and z is the axis of the cylinder which is the same as x3 axis. Cartesian coordinatex1 andx2 are in a plane normal to the z axis. Here,q1 =r, q2 =θ,and q3 =z.
The Jacobian matrix of transformation (63) is given by
Jq(x) =
cosθ sinθ 0 −rsinθ rcosθ 0
0 0 1
. (64)
Starting with equations (49), (63), and (64) we find that the cylindrical coordinates system is orthogonal and
ds1 =dr, ds2 =r dθ, ds3 =dz. (65)
Let the components of a vectorFbeFr, Fθ, Fzin ther, θ,andz directions, respectively. Following (58), (59), and (62), we have:
∇·F= 1
r
∂(r Fr)
∂r +
∂Fθ
∂θ +r ∂Fz
∂z
; (66)
∇×F =
1
r ∂Fz
∂θ −
∂Fθ
∂z
er+
+
∂Fr
∂z −
∂Fz
∂r
eθ+ 1
r
∂(rFθ)
∂r −
∂Fr
∂θ
ez,
(67)
whereeris the unit vector of the projected radius vector on the plane (x1, x2), eθ is a unit vector tangent in the point (r, θ, z) to the circle of r radius and the centre in the point (0,0, z), belonging to the plane parallel with (x1, x2) passing throught this point, andez is the unit vector of x3 axis. These three unit vectors stand fore1,e2,and e3 in equation (57),and the set {er,eθ,ez} forms a right–handed orthogonal basis in the three–dimensional space.
Ifu =urer+uθeθ+uzez is the displacement vector refered to cylindrical coordinates system, then the dilatation ∇u= Θ, the gradient of Θ,and the rotation vector Ψ=∇×u are given by
∇·u= Θ = 1
r
∂(r ur)
∂r +
∂uθ
∂θ +r ∂uz
∂z
∇Θ = ∂Θ
∂rer+
1
r ∂Θ
∂θeθ+ ∂Θ
∂zez, (69)
Ψ=∇×u =
1 r ∂uz ∂θ − ∂uθ ∂z er+ + ∂ur ∂z − ∂uz ∂r
eθ+ 1
r
∂(ruθ)
∂r −
∂ur
∂θ
ez.
(70)
From (70) we derive that the components of the vector Ψare
Ψr = 1
r ∂uz
∂θ −
∂uθ
∂z
Ψθ = ∂ur
∂z − ∂uz
∂r ,
Ψθ = 1
r
∂(ruθ)
∂r − ∂ur ∂θ . (71)
Our aim is to use the metric coefficients for the cylindrical coordinate system along with the above expressions for the vector and scalar functions to derive the Navier equations in the form given by equation (25) into cylindrical coordinates.
11. The Navier Equations in Cylindrical Coordinates
Using equations (68),(69),(70),and (59),whereF=Ψ,the Navier equations (25), divided by ρ, become in extended form
c2 L ∂Θ
∂r − c
2 T 1 r ∂Ψz ∂θ − ∂Ψθ ∂z = ∂ 2u r
∂t2 ,
c2 L
1
r ∂Θ
∂θ − c
2 T ∂Ψr ∂z − ∂Ψz ∂r = ∂ 2u θ
∂t2 ,
c2 L
∂Θ
∂z − c
2 T
1
r
∂(rΨθ)
∂r − ∂Ψr ∂θ = ∂ 2u z
∂t2 .
(72)
For longitudinal waves without rotation, the rotation vector Ψ vanishes and equation (72) reduces to the wave equation for the dilatation Θ. For transverse waves, Θ vanishes and equation (72) becomes the wave equation for Ψ.
We now investigate the propagation of radially symmetric waves in a solid infinite cylinder of radius a. Some authors ([?], [?] etc) take the following approach. They solve the Navier equations (72) for the dilatation and ro-tation components. From them they obtain the displacement and roro-tation. Then they apply the results to an infinite cylinder with free–surface boundary conditions.
The approach to be used here is to take advantage of the fact that an elastic medium has no internal friction, so that scalar and vector potentials exist. We shall make use of this fact by solving the wave equations for these potentials in cylindrical coordinates, and from these solutions we obtain the displacement vector, the stress components, etc.
Radially symmetric waves in a solid cylinder are symmetric about the cylinder axisz,so that the angular displacement componentuθ vanishes and the other components do not depend on the angle θ. Each particle of the cylinder oscillates in the (r, z) plane. It turns out that the rotation vector Ψ = ∇ ×u has a nonzero angular component that is independent of θ.
We consider an infinite train of time–harmonic waves of a single frequency along a solid infinite cylinder such that the displacement is a simple harmonic function of z. The cylindrical coordinates are (r, θ, z), where where θ is the angle that the radius vector r makes with the x axis. The x, y coordinates are in a plane normal to the cylinder axis z. The displacement vector is u = (ur,0, uz), where the components are functions of (r, z, t). We see that the rotation vector has the form ∇ ×u = (0,∂ur
∂z −
∂uz
∂r ,0) = (0,Ψθ,0),
where Ψθ is also a function of (r, z, t).
In our axisymmetric case, the relation between the components of the displacement vector to the scalar and vector potentials (33) become
ur =
∂φ ∂r −
∂ψ ∂z,
uz =
∂φ ∂z +
1
r ∂ ∂r
r∂ψ
∂r
= ∂φ
∂z + ∂2ψ
∂r2 + 1
r ∂ψ
∂r.
(73)
coordinates become
c2 L
∂2φ
∂r2 + 1
r ∂φ ∂r +
∂2φ
∂z2
= ∂
2φ
∂t2
c2 T
∂2ψ
∂r2 + 1
r ∂ψ
∂r + ∂2ψ
∂z2
= ∂ 2ψ
∂t2 .
(74)
The only nonzero components of the stress tensor are
τrr = λ
ur
r +
∂ur
∂r +
∂uz
∂z
+ 2µ∂ur ∂r ,
τrz = µ
∂ur
∂z +
∂uz
∂r
,
(75)
since the cylinder is axisymmetric so that the angular components of the stress tensor vanish.
The boundary conditions are on a free surface, which means that on that on the surface r=a we have
τrr = 0, τzz = 0, at r=a. (76)
To solve for the potentials we take time harmonic solutions. Furthermore, since the wave front propagates in thez direction, we assume the potentials are harmonic in z. Therefore we can separate the r−dependent parts of the potentials and write them in the form
φ =AF(r)ei(kz±ωt)
, ψ =BG(r)ei(kz±ωt)
, (77)
where A and B are constants and F and G are functions of r to be deter-mined. Note that the same phase kz±ωt is used for the scalar and vector potentials. At any instant t, potentials φ and ψ are periodic functions of z
with wavelength Λ, where Λ = 2π/k. The quantity k = 2π/Λ, which counts the number of wavelength over 2π is termed the wavenumber. At any po-sition the potentials φ and ψ are time–harmonic with time period T, where
T = 2π/ω. The circular frequency ω is given by ω =k·c.
Inserting equations (77) into the appropriate wave equations (74) yields
d2F
dr2 + 1
r dF
dr +β
2F = 0,
d2G
dr2 + 1
r dG
dr +γ
2G = 0,
where
β2 = ω 2
c2 L
−k2, γ2 = ω2
c2 T
−k2. (79) The solutions of equation (78) that has no singularity atr= 0 is Bessel’s functions of order zero, so that the potentials become, for progressing wave fronts,
φ=AJ0(βr)ei(kz−ωt)
, ψ =BJ0(γr)ei(kz−ωt) (80) Using (80), equation (73) for the displacement components become
ur =
A d
drJ0(βr) +ikB d
drJ0(γr)
ei(kz−ωt)
uz = [ikAJ0(βr) +γ2BJ0(γr)]ei(kz
−ωt) .
(81)
Starting with (69) and (81) we derive that the divergence of the displace-ment vector in the case of radially symmetric waves is given by
Θ =∇·u=−ω 2
c2 L
AJ0(βr)ei(kz−ωt)
. (82)
Using now the constitutive equations (75), the expressions of ur and uθ given in (81), and equation (82) we find that the components τrr, τrz of the stress tensor are
τrr =
2µd
2
dr2J0(βr)−λ
ω2
c2 L
J0(βr)
A+ 2µikBd
2
dr2J0(γr)
ei(kz−ωt) ,
τrz = µ
2ikAd
drJ0(βr) +
ω2
c2 L
−2k2
Bd
drJ0(γr)
ei(kz−ωt) .
(83) To satisfy the boundary conditions (76) we must to have
2µd
2
dr2J0(βr)
r=a−λ
ω2
c2 L
J0(βa)
A + 2µikd
2
dr2J0(γr)
r=aB = 0, 2ikd
drJ0(βr)
r=aA +
ω2
c2 T
−2k2
d
drJ0(γr)
The system (84) is a set of two algebraic homogeneous equations for the constantsAandB.Therefore, for nontrivial solutions thwe determinant must equal zero, yielding
2µd
2
dr2J0(βr)
r=a−λ
ω2
c2 L
J0(βa) 2µikd
2
dr2J0(γr)
r=a
2ikd
drJ0(βr) r=a ω2 c2 T
−2k2
d
drJ0(γr) r=a . (85)
This is the period equation. It is difficult to discuss this equation in its general form, other than to perform numerical analyses for special cases. But if we have a thin cylinder, than the radiusais small, thereforeβaandγa will be small enough to neglect fourth–order terms. This is seen by expanding the Bessel function
J0(x) = 1− 1 2x
2+ 1 64x
4
− · · ·, (86) wherexstands forβaorγa.Using equation (86) in the expansion of equation (85) and keeping only the quadratic terms, we get the required approxima-tion for the period equaapproxima-tion. From this we obtain an approximaapproxima-tion for the longitudinal wave speed:
cL =
ω k = s E ρ
1− 1 4σ
2k2a2, (87)
whereE is the Young’s modulus and σ is the Poisson’s ratio of the cylinder. 13. Waves Propagated over the Surface of an Elastic Body
Lord Rayleigh investigated a type of surface wave runing along the planar interface between air and an isotropic elastic solid in which the amplitude of the wave damps off exponentially as it penetrates the solid. It was anticipated by Lord Rayleigh that solutions of this type might approximate the behaviour of seismic waves observed during earthquakes [Proceedings of the London Mathematical Society, vol. 17 (1887), orScientific Papers, vol. 2, p. 441].
in its upper plane y > 0 and an isotropic elastic solid in the lower plane
y <0.We assume that monochromatic progressing waves (single frequency) are propagated in the positivex direction as a result of forces applied in the solid at some distance from the surface (for example, the forces that pro-duce earthquakes). Since the nature of the disturbing force is not specified, there are infinitely many solutions to these wave equations. However, using Rayleigh’s approach we obtain solutions that are exponentially damped.
We assume that all the dependent variables are functions of (x, y, t) and that the displacement vectoru = (u, v,0).The rotation vectorΨhas the form Ψ= (0,0,Ψ),where Ψ = ∂v
∂x− ∂u
∂y.The two–dimensional wave equations for
the scalat potential φ and the component Ψ of the vector potential Ψ are
c2 L
∂2φ ∂x2 +
∂2φ
∂y2
= ∂
2φ
∂t2,
c2 T
∂2Ψ ∂x2 +
∂2Ψ
∂y2
= ∂ 2Ψ
∂t2 .
(88)
The boundary conditions at y = 0 are of type of a free surface, so that the shear stress τxy and the normal stress τyy vanish on the x axis. In this case, Hooke’s law gives us
τxy = 2µεxy, τyy =λΘ + 2µεyy, Θ =εxx+εyy, (89) for two–dimensional stress and strain. The components u, v, and w of dis-placement vector u become
u= ∂φ
∂x + ∂Ψ
∂y, v = ∂φ ∂y −
∂Ψ
∂x, w = 0. (90)
The strain tensor has the nonzero components
εxx =
∂u ∂x =
∂2φ
∂x2 +
∂2Ψ
∂x∂y,
εxy = 1 2
∂v
∂x + ∂u ∂y
= ∂ 2φ
∂x∂y +
1 2
− ∂ 2Ψ
∂x2 +
∂2Ψ
∂y2
,
εyy =
∂v ∂y =
∂2φ
∂y2 −
∂2Ψ
∂y∂x.
Inserting equations (91) into (89) yields the boundary conditions for the components of the stress tensor at y= 0 :
2 ∂ 2φ
∂x∂y(x,0)− ∂2Ψ
∂x2(x,0) +
∂2Ψ
∂y2(x,0) = 0,
λ∇2φ(x,0) + 2µ∂2φ
∂y2(x,0)−
∂2Ψ
∂y∂x(x,0)
= 0.
(92)
The problem is to find time–harmonic solutions to (88) that exponen-tially decay with y, are progressing waves in the xdirection, and satisfy the boundary conditions (92). To this end we take the potentials in the form
φ(x, y) =Ae−ay
eik(x−ct)
, Ψ(x, y) = Be−by
eik(x−ct)
, a >0, b >0, (93) where A is the amplitude of the scalar potential, B is the amplitude of the vector potential, and a and b are the decay constants, which are determined from the wave equations. Note that the wave number k and the frequencyω
are the same for each potential. Inserting equations (93) into (88) yields
a2−k2
1− c
cL
2
= 0,
b2−k2
1− c
cT
2
= 0.
(94)
The wave velocity c is not equal to cL or cT. However, if a = 0 then
c=cL, or ifb = 0 then c=cT. But neithera nor b can vanish.
To satisfy the boundary conditions we insert equation (93) into equation (92) and obtain
−2iakA+ (b2 +k2)B = 0,
[2µa2+λ(a2−k2)]A+ 2iµbkB = 0. (95) Equations (95) are a pair of homogeneous algebraic equations for the complex constants A and B. As usual, we set the determinant equal to zero in order to have nontrivial solutions forA and B. We get
4µabk2−(b2+k2)
2µa2 +λ(a2 −k2)
Eliminating a and b from equation (96) by appealing to (94) we derive that equation (96) becomes a cubic in (c/cT)2,which can be put in the form
s3−8s2+ 8(3−2r)s−16(1−r) = 0, (97) where
s=
c cT
2
, r=
cT
cL
2 .
Using the approximationλ=µ(Poisson’s condition), we getr = 1/3 and equation (97) becomes
(s−4)(3s2−12s+ 8) = 0. (98) The roots of equation (98) are
s1 = 4, s2 = 2 + 2√3
3 , s3 = 2− 2√3
3 .
It is easily seen that the only root that yields positive values for a and
b is s= 2−2√3/3. We thereby obtain the following relationship between c
and cT :
c=√scT = 0.9194cT.
For the case of an incompressible body we have Θ = 0. This gives r= 0, so that the velocity of a Rayleigh wave becomes
c=√scT = 0.9553cT.
We have seen that in either case cis slightly less than the velocity of an equivoluminal wave.
We note that cis independent of frequency. This means that there is no dispersion, that is the wave shape is maintained. Having determined c in terms ofcT andcL,we can then calculate the decay constantsa andb,which determine the rate at which the potentials attenuate with depth (note that
y is positive downward). From the definition of the wave number k, both a
Seismographic signatures often depict waves similar in structure to Rayleigh waves. However, seismograph records of distant eartquakes indicate disper-sion, which means dependence of c on ω. This arises mainly because of the inhomogeneity of the earth, and also because of the viscoelastic properties of the earth.
14. Special cases
As we know, the vector Navier equation (24) is satisfied with a vector of the form
u=∇φ+∇×Ψ, (99)
where the potentials φ and Ψ have to satisfy the wave equations (suppose that F=0)
c2L∇2φ = ∂2φ
∂t2, c 2
T∇2Ψ=
∂2Ψ
∂t2 . (100)
A class of particular solutions of the Navier equation (24) can be generated by seeking solutions of (100) of the form
φ=Ae−ay+i(x−ωt)
, Ψ=e2e−by+i(x−ωt)
, (101)
where A, e2, a > 0, and b > 0 are constants. It is seen that in this case the displacement vector (99) has the form u = (u, v,0), and u and v are independent of z.The functions φ and Ψ in (101) are solutions of (100) if
a=
s
1 + ω 2
c2 L
, b =
s
1 + ω 2
c2 T
.
Since we have a two–dimensional problem it results that only the third component of the vector B is nonzero. Putting B3 =B, we follow to deter-mine the constants A and B from adequate boundary conditions.
When the gradient and curl are expressed in cylindrical coordinates (r, θ, z),
then for axially symmetric problems,
ur=
∂φ ∂r −
∂Ψ
∂z , uθ = 0, uz = ∂φ ∂z +
∂Ψ
∂r +
1
rΨ, (102)
and the functionsφand Ψ must to satisfy the following differential equations
c2L∇2φ = ∂2φ
∂t2, c 2
T∇2Ψ =
∂2Ψ
where∇2 is the Laplace’s operator in cylindrical coordinates which in axially symmetric case has the form
∇2 = ∂ 2
∂r2 + 1
r ∂ ∂r +
∂2
∂z2.
Particular solutions of equations (103) may be seeking of the form
φ=F(r)ei(z−ωt)
, Ψ =G(r)ei(z−ωt) ,
where F and Gare solutions of the following ordinary differential equations
F′′
(r) + 1
rF(r) +
ω2
c2 L
−1
F(r) = 0,
G′′
(r) + 1
rG(r) +
ω2
c2 T
−1
G(r) = 0.
(104)
The solutions of equations (104) may be expressed with Bessel’s function of order zero
F(r) = J0(β1r), G(r) = J0(γ1r), (105) where
β1 =
s ω2
c2 L
−1, γ1 =
s ω2
c2 T
−1.
If we have to solve a boundary value problem in axially symmetric case, then the functions φ and Ψ must to be of the form
φ=AF(r)ei(z−ωt)
, Ψ =BG(r)ei(z−ωt)
, (106)
whereF(r) andG(r) are given in equation (105),andAandB are constants which will be determined from boundary conditions.
References
[1] Love, A. E. H., A Treatise on the Mathematical Theory of Elasticity. Dover, 1944.
[2] Davis, J. L., Wave Propagation in Solids and Fluids. Springer–Verlag, 1988.
[4] Achenbach, J. D., Wave Propagation in Elastic Solids. North–Holland Publishing Company, 1973.
[5] Magnus, W., Oberhettinger, F., Soni, R. P., Formulas and Theoremas for the Special Functions of Mathematical Physics. Springer–Verlag, 1966.
Author: Ion Cr˘aciun
Department of Mathematics Technical University of Iasi
